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MHT CET 2026 (14 May, Shift 2) · Biot-Savart Law & Circular Coils

Biot-Savart Law & Circular Coils question from MHT CET 2026 (14 May, Shift 2)

Q12MHT CET 2026 · 14 MAY · SHIFT 2Medium

Biot-Savart Law & Circular Coils

A particle carrying a charge equal to 100100 times the charge of an electron is rotating one rotation per second in a circular path of radius 0.80.8 m. The value of magnetic field produced at the centre will be (μ0\mu_0 = permeability of free space)

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Answer: (b)

  1. Equivalent current I=qf=100×1.6×10−19×1=1.6×10−17 AI = qf = 100 \times 1.6 \times 10^{-19} \times 1 = 1.6 \times 10^{-17}\text{ A}.
  2. B=μ0I2r=μ0×1.6×10−171.6=μ0×10−17B = \frac{\mu_0 I}{2r} = \frac{\mu_0 \times 1.6 \times 10^{-17}}{1.6} = \mu_0 \times 10^{-17}
  3. So the answer is (b) μ010−17\mu_0 10^{-17}.
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